Let $\vec{x}$ be a vector in the plane containing vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$. If the vector $\vec{x}$ is perpendicular to $(3 \hat{i}+2 \hat{j}-\hat{k})$ and its projection on $\vec{a}$ is $\frac{17 \sqrt{6}}{2}$, then the value of $|\vec{x}|^{2}$ is equal to ____.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Vector Algebra

Let

x

be a vector in the plane containing vectors

a = 2 \hat{i} - \hat{j} + \hat{k}

and

b = \hat{i} + 2 \hat{j} - \hat{k}

. If the vector

x

is perpendicular to

(3 \hat{i} + 2 \hat{j} - \hat{k})

and its projection on

a

\frac{17 6}{2}

, then the value of

∣ x ∣^{2}

is equal to ____.

Answer: 486

Since

x

lies in the plane of

a

and

b

, write

x = λ a + μ b = (2 λ + μ) \hat{i} + (- λ + 2 μ) \hat{j} + (λ - μ) \hat{k}

Perpendicularity to

3 \hat{i} + 2 \hat{j} - \hat{k}

gives

3 (2 λ + μ) + 2 (- λ + 2 μ) - (λ - μ) = 3 λ + 8 μ = 0

With

∣ a ∣ = 6

, the projection condition gives

x \cdot a = \frac{17 6}{2} \cdot 6 = 51

. Using

a \cdot a = 6

and

a \cdot b = - 1

6 λ - μ = 51

Solving the two equations gives

λ = 8

and

μ = - 3

, so

x = 13 \hat{i} - 14 \hat{j} + 11 \hat{k}

and

∣ x ∣^{2} = 169 + 196 + 121 = 486

∣ x ∣^{2} = 486

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer